ANALYSIS OF PRESTRi^ooED AND REINFORCED
CONCRETE FOLDED PLATES
by
RAMESH GAMI
B. S., Bombay University, 1958
A MASTER'S REPORT
submitted in partial fulfillment of the
requirements for the degree
MASTER OF SCIENCE
Department of Civil Engineering
KANSAS STATE UNIVERSITY
Manhattan, Kansas
1963
Approved by:
'J/.yn,
ajor Professor
k f ii
TABLE OF CONTENTS
r>->e.i+s Page
SYNOPSIS 1
INTRODUCTION 2
DEFINITIONS 4
STATEMENT OF PROBLEM 6
ANALYSIS OF FOLDED PLATES CONSIDERING JOINT DISPLACE-
MENT 7
ANALYSIS 8
Primary Stresses 8
Secondary Stresses 20
ANALYSIS OF POST SSB8X0X5S rRESTREoSED COHCBEH FOLDED
PLATE 38
CONCLUSIONS 55
ACKNOWLEDGMENT 57
APPENDIX I - EXPLANATION OF TERMS 58
APPENDIX II - BIBTIOGRAPHT 60
ANALYSIS OF PRESTRESSED AND
REINFORCED CONCRETE
FOLDED PLATES
by
RAMESH GAUI 1
SYNOPSIS
The simplified methods of analysis for long span pre-
stressed and reinforced concrete folded plate structures are
presented herein.
A reinforced concrete folded plate is analyzed by resolv-
ing the applied load into two directions, one vertical and the
other parallel to the plate on which it acts. The vertical
load will induce "slab" action and the parallel load will in-
duce "plate" action. Slab action can be analyzed by assuming
non-yielding supports and then applying a correction for de-
flection, from which total plate loads are found. "Plate" ac-
tion causes a deflection of the longitudinal edge and intro-
duces longitudinal stresses. To satisfy the compatibility
condition, a stress distribution is done. Moments are com-
puted due to relative displacements. From these moments,
plate loads and longitudinal stresses are calculated.
Graduate student, Department of Civil xungineering,
Kansas otate University, Manhattan, Kansas.
To analyze the prestressed plate, the stress distribution
along the longitudinal axis is analyzed in the same manner as
for any homogenous beam of rectangular section.
Transverse bending in successive plates is analyzed by
the conventional moment distribution method utilized for con-
tinuous structures.
A typical problem is outlined and the simplified method
of analysis is discussed in this report.
INTRODUCTION
Folded plates, or as they are sometimes called, prismatic
shells or hipped plates, provide a useful and economical meth-
od of construction for roof and floor systems in a wide variety
of structures. They are competitive with other construction
methods for short spans and have proven exceptionally economi-
cal where relatively large spans are needed as for auditoriums,
gymnasiums, industrial buildings, hangers, department stores
and parking garages. The folded plate shape of roof structure
has come into wide usage because of its low cost of construc-
tion for long span, high load carrying capacity, rigidity, and
aesthetic interest. Selection of concrete for the shell mate-
rial furnishes a high degree of fire resistivity, ease of mold-
ing to the desired alinement and profile, a great degree of
permanence, and low construction and maintenance costs.
Folded plate floor or roof construction consists of a
series of repeated units, each of which is formed by tv/o or
more flat plates intersecting at an angle. The plates act as
a continuous slab transversely and as beams in their own planes.
Their structural behaviour resembles that of shells. In fact,
a cylindrical shell can be thought of as a folded plate in lim-
it. The structural action of a folded plate consists of trans-
verse "slab action" by which the loads are carried to the
joints and longitudinal "plate action" by which the loads are
finally transmitted to the transverses. Because of its great
depth and small thickness, each plate offers considerable re-
sistance to bending in its own plane. This "plate action" ex-
plains the remarkable rigidity of folded plate construction.
Folded plates have certain advantages over shells. These
advantages are:
(1) The shuttering required is relatively simpler as
it involves only straight planks.
(2) Shuttering can be stripped at the end of seven days,
if not earlier, because of their greater rigidity;
this results in quicker turnover which, in turn,
cuts down construction time.
(3) The design involves only simple calculations which
do not call for a knowledge of higher mathematics.
(4) Movable formwork can be employed for their construc-
tion with greater ease than with cylindrical shells.
(5) Simple rectangular diaphragms take the place of
complicated transverses required for shells.
(6) .?heir light reflecting geometry and pleasing out-
lines make then comparable with shells in their aes-
thetic appeal.
Interest in and use of this type of roof has increased
considerably in this country. Considerable additions have
been made to our analytical and experimental knowledge in the
last decade. The purpose of this report is to solve typical
problems and discuss procedures which may be employed for the
analysis of a single-span folded plate structure.
DEFINITIONS
The following definitions are used as a basis for the
discussion in this report:
(1) A plate is an individual planar element of the
structure.
(2) The length of a plate is the dimension between trans-
verse supports. (Fig. 1, "L")
(3) The width of a plate is the transverse dimention be-
tween longitudinal edges. (Fig. 1, "W")
(4) The height of the structure is the vertical dimension
of the upper and lower extremes of a transverse cross
section. (Fig. 1, "h")
FIG. 1. - END PORTION OF A FOLDED PLATE ROOF
Support
Support
FIG. 2. - TYPICAL CROSS SECTION THROUGH MULTIPLE FOLDS
STATEMENT OF PROBLEM
Consider a simple span of 60 feet in which we shall arbit-
rarily assume a column spacing along the ends of this longer
span of 23 feet. The slope of each plate will be assumed to be
3 1/2 on 12, vertical to horizontal. (Fig. 2.) A trial analy-
sis will indicate this corrugated configuration to be unsuit-
able for conventionally reinforced concrete due to the very
small area of concrete available at the ridge, causing high
compressive stresses and a very high percentage of compressive
steel reinforcement. This can be primarily attributed to the
shallow ratio of depth from ridge to valley for the long span
of 60 feet and the absence of sufficient concrete area in the
ridge.
Several remedies for this condition become readily appar-
ent:
(1) Steepen the slope of the plate and add a third plate
horizontally at the ridge in order to provide suffi-
cient concrete area to reduce the high compressive
stresses.
(2) Steepen the slope of the plates in order to provide
a deeper section from ridge to valley.
(3) Provide a combination of each of those stated in (1)
and (2).
(4-) Prestress the structure in order to utilize all of
the concrete area as homogenous section.
This report shall advance methods of analysis for design
of folded plates by the above mentioned procedures (2) and (4).
(a) To steepen the sloj^e of the plates in order to pro-
vide a deeper section from ridge to valley. The folded plate
is analyzed by considering the effects of relative displace-
ments of the longitudinal edges.
(b) To prestress the structure in order to utilize all
of the concrete area as homogenous section.
ANALYSIS OF FOLDED PLATES CONSIDERING
JOINT DISPLACEMENT
Assumptions
The following general assumptions are made in analyzing a
folded plate structure:
1. The material is homogenous, uncracked and elastic.
2. Longitudinal edge Joints are fully monolithic and con-
tinuous; there is no relative rotation or translation
of two adjoining plates at their common boundary.
3. The principal of superposition holds, that is, the
structure may be analyzed separately for the effects
of its redundants and various external loadings and
the results combined algebraically.
4. Individual plates possess negligible torsional resis-
tance and torsional stresses due to twisting of the
plates can be neglected.
5« The function of the supporting diaphragms or bents is
to supply the end reactions for the plate action and
8
for the longitudinal slab action. 1'hey are assumed
incapable of providing restraint against rotation of
the ends of the plates in their own planes, but may
provide some restraint for longitudinal slab bending.
6. Longitudinal strain due to plate action varies linear-
ly across the width of each plate (plane section re-
mains plane). The rate of change of strain with re-
spect to width ordinarily will differ from plate to
plate from which it can be inferred that there will
be some relative displacement of the joints of a cross
section.
7. Longitudinal slab action can be neglected; that is,
slab bending carries the load applied to the surface
of a plate to the longitudinal edges only, as in a one
way slab.
ANALYSIS
Primary Stresses
The folded plate for the structure, as discussed in
"Statement of Problem", is deepened at ridge and valley as
shown in the line diagram, Fig. 3»
FIG. 3. - LIKE DIAGRAM OF FOLDED PLATE STRUCTURES.
The data for this sample solution are as follows:
Slab thickness = 4 1/2"
Span between transverses * . . . 60' 0"
Column spacing = 23' 0"
Loading:
Roofing 5 psf
Snow load, insulation, and accoustatics ... 30 psf
The loads on the inclined surfaces are as follows:
For Plate AB (Fig. 4)
Snow Load:
FIG. 4. - PLATE AB.
10
cos*
£732
= 0.94-9
sin* = 2 0.317
5732
Snow load on inclined surface
= 30 x cos* = 30 x 0.949 = 29. 4-7 lbs/sq. in.
Dead Load :
Hoofing material = 5*00 lbs/sq. in.
Slab 4.5 x 150 * 55.33 lbs/sq. in.
12
Total Load :
Sum of above three
= 29.^7 ♦ 5.00 + 55.33
= 89.80 say 90.00 lbs/sq. in,
Components of Total Load : ( Fig . 5 • )
This load can be resolved into normal and tangential
components, Wn and Wt.
FIG. 5. - NORMAL AND TANGENTIAL COMPONENTS ON PLATE AB.
11
Wn ■ W x cos <x
= 90 x 0.949
= 85.41 lbs/sq. in.
Wt = W x sin <x
= 90 x 0.317
= 28.53 lbs/sq. in.
For Plate BC (Jig. 6.)
fc-j —
FIG. 6. - PLATE BC.
Snow Load:
Dead Load:
« as x 11.5
12.92
■ 27.62 lbs/sq. in.
roofing ■ 5.00 lbs/sq. in.
Dead load of slab m 55. 33 lbs/sq. in.
Total Load :
= 87.95 lbs/sq. ft.
Components of Total Load :
Normal component Vn ■ 87.95 x 11.5
12.95
= 77.5 lbs/sq. It.
Tangential component = Vt = 87.95 x 6
12.92
= 40.60 lbs/sq. ft.
12
The load on DE is by observation the same as the load on £C,
that is:
w = 77.5 lbs./sq. ft. say 78 lbs./sq. ft.
W t = 40.6 lbc/s^. ft. say 41 lbs./sq. ft.
For the purpose of analysis, assume fictitious supports at
joints B, G, D, and calculate the Lioments and reactions. It is
assumed that the relative displacement of joints is not present.
A transverse strip one foot wide is considered, treating it as
continuous slab supported at the joints by non-yielding sup-
ports. The Boment distribution for this condition is performed
in Fig* 7*
86#/i
't
76#/ft
8(
s#
/t
t
'1
'
1 ■
\
•
■ '
i
f )
' 1
r
'
\
' '
'
r ^
1' ,r v | | , V I 1 j
■ v i 1 i> v ■' ' » 1
A
i
E
r *
c
fD
E
A E
c
£
Dist. F.
1 0.5
0.5 1
FEM +1725
-1090 +1090
-1090 +1090
3,0.
-635 -317.5
+317.5 +625
final n +1725
-1725 +772.5
-772.5 +1725
-1725
FIG. 7. - SLAB MOMENT DUE TO EXTERNAL LOAD.
Reactions are as follows:
Total reaction at B ■ 1124 lbs.
Total reaction at C s 368 lbs.
Total reaction at D = 1124 lbs.
13
These reactions were obtained on the assumption that fic-
titious supports exist at B, C % and D and hence, on removal,
they give e jual and opposite reactions. These reactions are
placed on the plates as shown in fig* 8,
FIG. 8. - PLATE SHOWING REACTIONS DUE TO SLAB ACTION.
These plate loads due to the above reactions can be ob-
tained as follows (see Fig. 9) J
tan © x - § - 0.3333
© x « 18° - 26»
tan © 2 * jj-r = 0.521
© 2 s 27° - 40'
Considering joint B, (see Fig. 10)
Comp. along; AB Comp. along BC 54-8
sin(90 + © 1 + 9 2 ) = sin 90 s sin (ISO - © x
*P
Comp. along AB . ^ 8 x sin <*> » & 1 * V
sin (180 - 57 - 9 2 )
= 5^8 x 0.96 = 526 lbs.
14
FIG. 9. - INCLINATION OF PLATES.
Comp. along BC ■
54-8 x sin 90
sin (180 - 57 - 9 2 )
548
~ 0.?1
= 770 lbs.
5^8#
FIG. 10. - COMPONENTS ALONG AB AND BC DUE TO REACTIONS
Similarly due to the load of 576 lbs. the components along
AB and BC can be obtained as follows:
Comp. alonp; AB Comp. alom BC 576
sin 90 = sin (90 * ^ 4 © 2 ; = sin (180 - 9 X - © 2 ;
Comp. along AB = s JPf 1 g Q 3 ^ Q . Qg)
* o?fr = 812 lbs#
15
576 x sin (90 f & 1 f © 2 )
Comp. along BC = sin (180 - ^ - © 2 )
= 576 cot (9« + w 2 )
- 576 x 0.96
= 553 lbs.
Considering joint C, seo Fig. 11.
FIG. 11. - COMPONENTS DUE TC EEACT10H8 A? C.
Comp. BC Coinp. along CD
sin 54° - 40' " sin 90°
Comp. along BC * 434 x ffiB g^L ~
sin 55°
^ x 0.82
= 297 lbs.
Comp. along CD = s in 55*- 20 '
434
" 0.82
» 528 lbs.
Due to 434 lbs. reaction
Comp. along BC - 528 lbs.
Conp. along CD = 297 lbs.
434
sin (180 f 2 9 2 )
40'
"20^
16
812
316
297
FIG. 12. - TOTAL FORCES AT THE JOINTS
The total forces at the joints are shown in Fig. 12.
The total load along AB can be calculated as follows
Load along plate AB =» 1358 * Wt x 6.32
= 1358 f 29 x 6.32
= 1521 lbs.
Similarly total load along plate BC
= 2149 f v/t x 12.92
= 2149 f 528
= 2677 Ids.
17
From these loadings we can find longitudinal stresses on
each plate. In the analyses one treats each plate as a beam
carrying loads and spanning between end diaphragms with no edge
shear along joints.
Plate AB (Fig. 13.)
1521#/ft
1 1 1 1 1 1 1 1 1
a'
so.o'
B
A
rf***-
B
1
1880#/in 2
1880#/*
FIG. 13. - STRESS DISTRIBUTION ALONG PLATE AB.
f
1
M
w x L
8
f
p
w x L x h
s
x 12 1
8 x 2 x b
x h 5
2 T 2
3 x w x L
4 x b x h^
3 x 1521 x 60 2 x 12
4 x 4.5 x 6.32 x 6.32 x 144
■ 1880 lbs/sq. in.
The stresses in top and bottom fiber are
« 1880 lbs/sq. in.
T t " t
f ju • - 1880 lbs/sq. in.
13
Piatt BC. (Fig. 14.)
2149#/ft
B
60.0'
V
4.5
m^M
FIG. 14. - STRESS DISTRIBUTION ALONG BC.
-9ia#/ n a
f *
. 2
X W Y. If
4 x b x h
x 2.49 x 60 x 60 x 12
x 4.5 x 12.92 x 12.^2
912 lbs/sq. in.
1860 ^An 3 -
6 ld&0#/in x
912 #/in
FIG. 15. - STRESSES AT JUNCTIONS OF PLATES
19
From the preceding analysis, it Is seen that the stresses
at the junctions of plates AB and BC at B are different as
shown in Fig. 15» but the compatibility condition requires them
to be equal. To satisfy the compatibility condition, a stress
distribution similar to the process of moment distribution is
done where,
Stiffness •<
1_
bh
Carry over factor ... = - #
Since the thickness of AB and BC are equal, the stiffness fac-
tors are as follows:
K ab = £732
= 0.160
if L
ab 12.92
= 0.078
The stress distribution is as follows:
A 'A
B % (
Z %
D g
E
DF
0.672
0.328 0.5
0.5 0.328
0.672
Stresses
+1880 -1880
-912 +912
+912 -912
-1880
+1880
CO.
-325 +650
-318 +159
+159 -318
+650
-325
Final
Stress
+1555 -1230
-230 +1071
+1071 -1230
-1230
+1555
Sign Convention:
Compression + V
e
Tention
- V.
e
20
Secondary stresses
Assume that plates AB, BC, CD and BE are not rigidly
jointed and solve for the free edge deflections. There will
be stresses due to rotations of those plates which do not have
a free edge. Considering a transverse strip one foot wide at
the middle of the structure
-5 x L 2 (f. - f K )
S *
48 x E x h
where f-. ■ stresses at left edge of plate
f. = stresses at right edge of plate.
The deflection of plate BC can be computed as follows:
f f b
S
-5 x L 2 (f . - f. )
be 48 x £ x h
5 x 60 x 60 1071 - (-1230) x 144
48 x £ x 12.92
5 x 5600 x 2300 x 144
= 48 x 12.92 x E
18 x 23 x 144 x lQ^
= 48 x 12.92 x E
be = E
S. , - 96.0jc 10
Similarly for AB
g - 5x 60 x 60 (-1230 - 155) x 144
max = 48 x E x 6.32
5 x 3600 x 2785 x 144
■ ♦ 48 x 6.32
S ab = ^ * 1Q 3 inches.
21
From these individual deflections, the composite deflec-
tion is found out "by graphical construction as shown in Figure
17. From Figure 17» we obtained the deflection of
170 x 10"^
BC ■ — i — 6 . This deflection is obtained on the assumption
that edges A, B, D, D, and E, are all free but actually this is
not the case and the deflection will be very, very small due to
the fixity of the Joints. Therefore, moments are calculated
which are produced by this relative displacement. In case of a
fixed end beam, moments induced at A and B by sinking of sup-
ports B will be - — *— , see Fig. 16.
FIG. 16. - MOMENT DUE TO DISPLACEMENT,
Considering edge C as the fixed end and end B as the end
free to rotate, the bending moment at C = • — *r- .
22
s A b " 2 ^¥^ ^ -" s
e BC - 9 -H^ »» -i*-
e BC : ^^^ ^ units
Scale: 1" = 60"
FIG. 17. - COMPOSITE DEFLECTION ALONG BC .
I =
b x t
I =
12
1x4
.?
12 5 x
12
1
19 x
12
M.
3 x £ x 170 x 10-
where 8 =
19 x 12 x E x 12.92 x 12.92
170 x lp5
E
L = 12.92 ft,
hi - 1360 lb. ft.
The calculation of shears is as follows
23
B
12-92'
&
'at© 12.92'
J
1360 #•**■
*b K c 12.92
= 105 lbs
The components of the loads in the direction parallel to
plates are
Plate load along AB a ^§j
= 148 lbs.
Plate load along BC a 105 x 0.96
= 101 lbs.
Plate load along BC = 105 x 0.^6
0.82
= 71 lbs.
24
Plate load along BC is therefore = 101 - 71
= 30 lbs.
Plate loads along AB and BC are, (see Fig, 18.)
Plate load along AB » 148 lbs.
Plate load along BG ■ 30 lbs.
FIG. 18. - PLATE LOADS ON AB AND BC,
From this the moment on plate AB can be obtained as follows:
B t E
/
With tension as (-) and compression as ( + )
R x 60 = 2/3 x 60 x 148 x 30 and
a
M Q = 2/3 x 148 x 30 x 30 x 5/8
= 55500 ft. lbs.
For plate BG the moment is as follows:
30#
go'
^L
25
M Q = 2/3 x 30 x 30 x $0 x 5/8
= 75 x 150
= 11250 ft. lbs.
M
Stress in plate AB » -x
„ b x h 2
Z * T —
Z « 4 * 3 x 6 »?2 2 x 144
108 x 40
,3
* 4320 in*
Stress in plate AB - ^°? 2 * 12
=» 154 psi
Stress in plate BC ■ 8
11250 x 12
Z
7 b x h 2
4.5 x 12.92 x 12.92 x 144
6
Z « 18000 in 5
Stress in plate BC = 112 ^ n ^ n 12
18000
- 7.5 psi
Stress condition for plates AB and BC are as shown in
Figure 19.
26
1.54- #/in
154- #/i^
^7-5 #/<rf
FIG. 19. - STRESSES IN AB AND BC.
Compatibility requires stresses at the junction to be
equal and hence stress distribution is carried out as follows:
A
3
C
0.672
0.328
-154
+154
-7.5
+7.5
+54.25
-109.5
+53
-26.50
-99.75
+45.5
+45.5
-19.0
cresses
CO.
Final Check
The maximum deflection is obtained once again from which
relative deflection is commuted by the graphical procedure as
in i ? ig. 21.
B ab = -5/48 x L 2 x (f t - f b )
= +5A8 x 60 x
60 (4^5 + 99.7 5)
E x 6.32
■ 12 ' 2 $ * 10? inches
be = -5/48 x 60 x 60
(43.5 + 19.0) x 144
12.92 x E
5/48 x 3600 x 64.5 x 144
12#92
- 2 -?° J 1Q ^ inches
27
Individual deflections are as shown in Fig* 20, from which
relative deflection at BC can be found out graphically as shown
in . 21. Lfta 21, s bc = ll '° | 1Q inches
We see that correction applied for a displacement of
1 ? Q 1 1Q causes itself a displacement of 11,Q * 1Q . The
11 x 10^
correction will have to apply for a displacement of *** w A
which will be in form of geometric series, such as
2 n
a, ar, ar , ar
Summation
1 f r
In the above case the series will be as shown below after
taking out common factor of *jr
170, 170 x £^» 170 x ( n *g^
170
Summation = , . 11.0
1 * T75
m 170 x 170
" 181
160
P^t^-K Q 160 X 10^
Correct S^ « 1
Actual S, m = Actual Longitudinal stress Actual Moment
oc - ■ f
Initial S, Initial Longitudinal stress Initial Moment
160.0
= 0.94
28
FIG. 20. - INDIVIDUAL DEFLECTION AT AB AND BC .
cj.-d = ^—y. in units
r, 2.70 x 10? .
b BC = — ' = m units
o 11.0 x 10 ?.
6-op = jg in units
Scale: 1" = 10"
FIG. 21. - RELATIVE DISPLACEMENT ALONG BC .
29
The longitudinal stresses at center line of structure due to
deflection are multiplied by 0.94- by which we obtain the actual
longitudinal stresses.
Longitudinal stresses at the center line of the structure
due to deflection are:
Initial stresses
Final stresses
Initial xo 94
-99.75
-95.75
^5.5
+42.8
45.5
+42.3
-19.0
-17.8
The total longitudinal stresses at center line are:
A
1
C
+1550.00
-93.75
-1230
+42.80
-1230
+42.80
+1071
-17.80
+1461.25
-1187.20
-1187.20
+1053.20
stresses due to load
(psi)
Stresses due to de-
flections in psi
Final stresses in psi
Final end moments :
A
3
_J
*
J
+1725
-1725
+77-50
-772.50
+1725
+1360
-1360
+1280
-1280
Final end moment due
to loading
'inal end moment due
to deflection with-
out correction
Final end moment due
to deflection with
correction
1360 x 0.94 * 1280
30
Moments at center of long n (60'),
Moments due to load
w x If
c 78 x 12.92 x 12.92
8
= 1620 lb. ft.
The moment diagram at center of long span is shown in
Fig. 22.
FIG. 22. - MOMENT AT CENTER OF LONG SPAN.
Moment at quarter point of long span (60') due to deflection
(Fig. 23):
The moment due to deflection is assumed to have parabolic
variation along the span of structure.
nT) ,. , Q . 1280 x 13 x 13
Ordinate ab = 30 x 30 —
= 32u units
Moment at quarter point of long span due to deflection =
1280— 320 = 960 ft. lb. (See Fig. 23.)
31
>^
"TC
-fi
O
.1
u
< *'
* X5'
15'
1 — ar
-w
O
oO
FIG. 23. - MOMENT AT QUARTER POINT DUE TO DEFLECTION.
Moment diagram at quarter point: (Fig. 24.)
Moment at C * 772 + 1280 - 320
* 1732 ft. lbs.
FIG. 24. - MOMENT DIAGRAM AT QUARTER POINT OF LONG SPAN (60')
Moment diagram at end of long span (60'): (?ig« 25.)
Moment due to deflection ■
FIG. 25 - MOMENT DIAGRAM AT END,
32
Shear force at center line of long span: (Fig. 26.)
86#/ft
IU1 ! 1 1 4 1111 1
2*
1725^
^ 78#/ft
B lll 11 1 11 11 i 1 1 1 1 ITT] c
2050#f<t
as.ss*
FIG. 26. - SHEAR FORCE AT CENTER LINE OF LONG SPAN.
For Span BC:
Reactions due to Loadings:
R B = R c = 78 x 2^2 = 505 lbs.
Reactions due to Moment:
R . R . 172^ -202?
K B - *C 12.92
* - 25.25 lbs.
For Span AB:
R B = 5^0 lbs.
Total shear force at
C = 505 - 25.25 ■ 479.75 lbs.
B = 505 - 25.25 + 5^0 = 1011.75 lbs.
33
Shear force at quarter point of long span: (Fig. 27 •)
86#/ft
a 1 1 1 i i a 1 i i i 1 b
78#/ft
<o4-0# ' 505 * 505#
FIG. 27. - SHEAR FORCE AT QUARTER POINT OF LONG SPAJN.
For sjjan BC:
Reactions due to loading
R B = R c = 505 lbs.
Reaction due to moment ■ negligible
For span AB:
R^ x 6.32 = 1725 + 86 x 6.32 x —^
R^ = 5-4-0 lbs.
Total shear force at
G = 505 lbs.
B » 505 + 54-0 = 1045 lbs.
Shear force at end of long span (60*): (Fig. 26.)
86#/ft
BPH 3 ^ilUUUI UUUlj C 772 Ft#
: V^ 505.00 * i
54-0* 73-60 #= 73. 6#
FIG. 28. - SHEAR FORCE AT EHD.
34
For span BC:
Reaction due to loading « 505 lbs,
.Reaction due to moment = 1725 - 772
12.92
= 73.6 lbs.
Total shear force at
G = 505 + 73.6 = 578.6 lbs.
B = 5^0 + 578.6 = 1118.6 lbs.
Final results may be summarized as follows:
Maximum longitudinal stresses (consider compression as
positive and tension as negative) :
,
B
G
f 1461. 50
psi
-1187.20
psi
+1053.20
psi
Maximum moments (moments in ft. lbs.):
B
C
D
Location
-1725
-2052
-1725
fe mid span
-1725
-1732
-1725
@ 1/4& span
-1725
-772
-1725
at end
35
B
6- 32
12-92'
t
B.M. Diagram
is plotted on
Tension Side.
PLATES AB AND BC.
479.75
SHEAR FORCE AT CENTER OF LONG SPAN.
7
2052
BENDING MOMENT AT CENTER OF LONG SPAN
FIG. 29. - BENDiNG MOMENT AND SHEAR FORCE DIAGRAMS
AT CENTER OF LONG SPAN.
B
<t
< 6-32' J
12.92'
w
PLATES AB AND BC .
36
505
SHEAR FORCE AT QUARTER POINT
1732
BENDING MOMENT AT QUARTER POINT.
FIG. 30. - SHEAR FORCE AND BENDING MOMENT AT QUARTER POINT
B
37
I-
6- W
->*-
12.92'
PLATES AB AND BC .
578.6
SHEAR FORCE AT END OF SPAN.
772
BENDING MOMENT AT END OF SPAN.
FIG. 31. - BENDING MOMENT AND SHEAR FORCE AT END OF SPAN,
t>*
. IS OP POSMJiM I0B20 FKEST8E88£I)
CONCRETE FOLDED PLATE
x'he stress distribution acting axiaily ^long the longitu-
dinal axis is the same for any homogeneous beam of rectangular
section. The top surface will be in compression, the bottom
in tension under gravity loads. Application of a sufficient
pre stressing force at a suitable distance below the center of
gravity of the section c ; in eliminate the tension stress.
The transverse direction of the plate is analyzed as a
continuous slab of length equal to the width of the plate and
supported at the folded lines of the ridges and valleys. These
spans being quite short thus require only nominal steel rein-
forcing as dictated by slab thickness and bending moment. Re-
ductions of the plate action deflections along the ridges, and
valleys, due to application of prestressing force, permit anal-
ysis of the transverse slab by moment distribution with a rea-
sonable degree of accuracy. Distribution bars are provided in
the longitudinal direction in an amount ordinarily utilized for
temperature steel.
The design of the structure is more easily understood with
the assumption of one individual leaf isolated from the others.
One may consider the plate oriented in its working position and
compute its sectional properties about the horizontal axis
"Long Span Prestressed Concrete Folded Plate Roofs", by
J. Brough and B. Stephens, Proc. Am. Soc. G. E. , Jan., I960,
p. 95.
39
extending through its center of gravity. Figure 32 indicates
the terms used in the following presentation.
FIG. 32. - SECTION PROPERTIES OF ONE FOLDED PLATE.
First a slab thickness is assumed and then the moment of
inertia of the cross section for the inclined position to the
vertical axis is computed. The moment of inertia of one plate
is derived as follows with a slab thickness of 4# inches, nor-
mal to the slope, and with the slope of the plate being 3»5
vertical to 12 horizontal.
Thus, in Fig. 32:
dA - tdx
y « tanc£
fe-fc*
dx
12 dy
-frr
♦H/2
-H/2
♦H/2
-H/2
y 2 t dx
_2+. 12 dv
40
U-l 1^2 fH/2
-H/2
T3~T
12 t
" " '■ ' .Li
3 3
12 t / H v
T3 ( 57 + ^
H' t
♦H/2
-H/2
H 5 t
where t is vertical thickness of the slab
i is the slope rise on 12 which is 3.5
H is height of valley to ridge center to center
t' is the •!«% thickness normal to slope
Thus t - |— .• Ui x
P P
12^ + 3.5
* 4,5/0.98
4.6875 in.
j . ft^p x (z.o.25) 5
I - 87300 in
The tensioning force required for zero stress in the bottom
under working load, at midspan is
I =
e + k.
41
Where F is the prestress force
e il the ecentricity of prestress force from the center
of gravity.
k is the kern point
The top fiber stress, at raidspar under workin- load is
f FH
ct ^ A c°b
Where f is the allowable concrete unit stress.
c
H is the Ytrtieal height of the section valley to ridge
A^ is the area of concrete,
c
The bottom fiber stress at initial condition at tr isfer is
f =* F i
cb A
A c
1 + e - U fx /F ±
k t
Where subscript i indicates the condition at transfer of pre-
stress.
Area of one plate = 4.6875 x 11. 5 x 12
= 647 in 2
r A 647 L ^
r r Htt m 40.25 + 4.6875 oo n.n u
r 2
k. * k, = •—- = 6.0 in.
t b c
Where k. is top kern distance
k. is bottom kern distance
r is the radius of gyration
42
If one assumes the center of gravity of the tendons as
being e' above bottom of section at midspan, the prestress
force eccentric arm is
e ■ c - e* = 16.4-7 i'-i.
Assume e' ■ 6.0 in.
The longitudinal bending moments acting on the section for
the dead load of the slab, the applied dead loads such as the
roofing, insulation and ceiling, and the live load for which
the structure shall be designed are computed as follows:
Dead load for slab
&*S 12 144 " ^ K? on inclined surface
Vertical load
2Z— a ZS- at S6 PS
cos<$> 079^ jK> '^
Total load per plate
11.25 x 56.25 = 646 PSF.
Additional applied dead load:
Roofing (five ply built up asphalt and felt) * 5*0 FSF
Concrete insulation and }4 in. acoustical plaster = 5»0 PSF
Total applied dead load:
10 x cos <J> ■ 9.6
Applied dead load per plate
9.5 x 11.5 = HI lbs.
Snow load per plate
25 x 11.5 ■ 287.5 lbs.
43
Moments can be computed as follows:
M^ slab - 0.6*6 X *jp • 291*0 ft-kip
2
M dl a PP lied - 0.111 x ^|- * 50.0 ft-kip
2
HL, * .2875 x S|- « 129.6 ft-kip
M tl " 47 °* 6 ft ~ ki P
The required prestre3sing force to furnish zero stress in
the bottom -\t mid span with total load moment is
M ti
470.6 x 12000 , n ft , . a
88 16.47 + 6.0 = 250 '° klps
■ e + K t 16.47
F o " $k * ** kipS
P * Total effective prestress after deducting losses.
F ■ Total prestress just after transfer, using say 4
cables, requires F = 250/4 «= 62.5 kips per cable.
F. ■ 73^5 kips per tendon.
Initial tensioning force required for each cable should
increase adequately for overcoming tendon friction and wobble
of conduit.
F H
Unit stress at mid span = F. = -r—g
c b
250 x 44.94 x 1000
= 647 x 22.47
= 770 vs i top comp.
= psi bottom comp.
44
Under the slab load only, the unit stresses at time of
transfer or prestress force will be
i • - K
¥
16.47 - 291. J
57o
x 2.54 x 1000
= 1155 psi comp.
Using allowable steel and concrete stresses specified by
the Joint Committee 323 ACI, ASCE for pres tressed concrete for
2
tensioned member.
Prestressing steel f 's « 240,000 psi
f 's yield = 210,000 psi
s
ultimate strength of steel
f ' at design load ■ 3750 psi
c
determine the area of prestress steel
A s = f~ s lffioOO * *»73* ■*• in "» sa y 1 -75 sq. in.
s
Check the unit shear stress:
V = hi | fiS « 33.3 k ips
Assume tendons placed in parabolic curve with c-k = e
» 22.47 - 6.0 - 16.47 in above bottom at ends
Joint Committee $23 ACI, AJ3CS, for crestrsssed concrete
for tensioned members.
45
FIG. 33. - LOCATION OF FRESTRESS FORCE ARRANGED IN
PARABOLIC CURVE.
Shear at cracking load in steel:
xr 4 x F x h
V s ' L
Where V is shear in steel
s
F is prestressing force after losses
L is length of span, center to center of supports; 61.5'
h is vertical distance below center of gravity of con-
crete section, 10.45 in.
v 4 x 286 x 10.45
V s * 61.5 x 12
* 142 kips
Shear force in concrete:
V Q
v„ = c^
C f
Where Q is statical moment about center of gravity of concrete
section
46
I is moment of inertia of section
b is horizontal projection of i>late width
19.1 x 23 60 loc; mm4
v c = 87300 x 43 " 125 pSi
The principal tensile 3 tress is gi;-en by
s t
v c 2 ♦ (t c /2f- - f 0/2
Where f is allowable concrete unit stress
c
f c ' ^T 1 = 386 psi
s t = ( >5 2 ♦ 193 2
« 37 psi, which is satisfactory, since the allowable
tensile stress is 110 psi.
Check shear at ultimate load;
u 13-
Where V is ultimate concrete shear
c
2V - v g = 66.6 - 14.2 * 52.4 kips
v is ultimate shear stress
52.4 x 2560
v u " 87300 x 4.5
* 351 psi
46 a
Ultimate principal tensile stress is given by
V
v u 2 +< f c/2> 2 * f c/2
551 2 +193 2 - 193
» 199 psi, which is satisfactory, since the allowable
tensile stress is 300 psi, thus design is satisfac-
tory.
Check section for cracking moment:
The factor of safety against cracking moment
* M/M g
Where M - M Q + M„
s c
M 8 = F (e + k t )
286 (16.47 ♦ 6.0)
12
= 514.10 ft. kip.
M F_.I 0.14 f • x I
r^ ■ c * c
c
%
« c = °- 14 f 2 ^° x x il 7?0 ° ■ w "• "*•
The factor of safety against cracking:
514.10 - 170 , y.
525 i# ^
The factor of safety against live load cracking:
M - M
; dl
M
11
47
Where H • M. ♦ ML ■ 684.10 ft. kip.
s c
^dl * s total dea( * load moment » 341.0 ft. kip.
M,, is live load moment . . . - 129.6 ft. kip.
The factor of safety against live load cracking:
684.10 - 341.0
= 120.6 —
= 2.65
The factor of safety considering ultimate moment:
The factor of safety for total load
"u
^n~ ~~ -m ultimate tensile force x ultimate lever arm
Where fi * pj
s
Ultimate tensile force ■ f ' . A a
c s
* 240,000 x 1.75 x 0.001
m 420 kips.
Ultimate lever arm can be computed as below: (See Fig. 34.)
Ultimate lever arm m 44.94 - 6.0 - distance of center of
concrete resistance below top
Center of concrete resistance if found as below: (See Fig. 34-.)
Equivalent width of top section « 12 * 4 *5 = 15.4-3 in.
Area of top triangle = ^* 6< ? * 1 ^' 4 ? « 36.2 sq.in.
48
FIO. 34. - ULTIMATE MOMENT STRESS CONDITION.
Additional compressive area of stem necessary is
m Total area under compression - area of top triangle
» 420 x 1000
5190
* 95. 3 s<i« in
- 36.2
Additional vertical depth of stem required is
■ fill? = 6 - 2 in -
Taking moment about top for finding center of concrete
resistance "C ''.
cr
c m ?6.2 x 2/3 x 4.6? jfcjj (6.2/2+4.6? )
^cr 131.5
■ b.5 in. below top
Ultimate lever arm is
« 44.94 - 6.0 - 6.5
* 32.44 in.
Ultimate moment
- " 20 i^ 2 - m . 113 ft. kip.
49
The factor of safety for total load
M s 11^8
m O
The factor of safety for live lord
- M ; M di
1138 - 341
* 129.6
= 6.15
Transverse bending in successive plates is analyzed by
the conventional moment distribution method utilized for con-
tinuous structures. The valley and ridge serve as support for
the slab. This being a short spsji, 11" - 6" f moments will be
no greater than would be encountered in a level slab with sup-
ports at the same frequency.
Analysis of a 1 ft. wide strip transverse to the main span:
W dl * 54 ♦ 10
= 64 lbs.
w\ • » 25 cos (J)
= 24.5 lbs.
V total load =64+24.5
= 88.5 lbs.
Assuming a strip of unit width of cross sectior of the
structure to act as a continuous one way slab on unyielding
50
supports, the ridge moments are determined by the moment dis-
tribution. Because of symmetry, the slab may be assumed as
rigidly fixed at ridge.
11.5*
A
B
1
5/7
V7
+1725
-1060
-665
+1060
-333
-1060
+1060
+143
+190
+95
+1725
-1725
+870
-870
+1155
FIG. 35. - RIDGE MOMENTS.
Ridge moments are computed in Figure 35*
Maximum bending moment at the center of span occurs when
that span and every other alternate span is loaded. Thus, bend-
ing moment at center due to uniformly distributed loading can be
^"Design of Folded Plates", by Eliahu Tranm, Proc. Am. Soc.
C.E., Vol. 85, Oct., 1959, p. 87.
51
obtained by the following formula:
riidsj;an positive bending moment ■ rr- (w, + V, /p) L
= }g x (64 ♦ 12.25) x 12 2
= 912 lb. ft.
Distribution bars are i>rovided in the longitudinal direction
in an amount ordinarily utilized for temperature steel. The
positive moment reinforcement, extending transversely, was
placed on top of the distribution steel. The prestressing ten-
dons were placed directly on top of the bottom steel and tied
to the bottom mat at the correct location.
Lfter all tendons were placed, the end bearing plates were
substantially anchored to the end form.
Reinforcing bars for slab negative moments at valleys and
ridges, stirrups, ties and grids for resistance of bursting of
concrete at the end anchorage were placed last.
Deflection:
The moment due to pres tress is
P x e = 294 x jo ■ W kip-ft for uniform loading
plus 294 x p ' * 256 kip-ft for parabolic prestress.
K dl slab = 29.10 ft-kip
M dl a P^ lied ■ 50.0 kip-ft.
n i;L = 150.6 ft-kip.
52
Upward deflection due to uniform prestress of 14-7 kip-ft:
o
W x IT
147 x 6Q 2 x 12 2 x 12000
8 x 3,500,000 x 87300
0.374 in.
Upward deflection due to parabolic prestress is
5 x V x L 2
* Sth —
5 x ?56 x 60 2 x 12 2 x 12000
= 3,500,000 x 87300
« 0.5^3 in.
Total instantaneous upward deflection due to prestress
- 0.374 + 0.543
* 0.917 in.
Deflection after losses
» 0.85 x 0.917
• 0.78 in.
Downward deflection due to slab load
5 x 291 x 60 2 x 12 2 x 12000
48 x 3,500,000 x 87300
= 0.617 in.
H«n04 immediate upward deflection at transfer is
0.917 - 0.617
- 0.300
53
Dead load deflection downward
30 x 0.617
* 391.0
« 0.162 in.
Net downward deflection after losses of prestress and effect
of creep is
» 0.300 - 0.162
« 0.138 in.
Instantaneous downward deflection due to live load moment is
130.6 x 0*617
291.0
- 0.275 in.
The resulting camber due to application of prestressing force
supports the plate in its geometrical position thus eliminating
critical secondary stresses caused by rotation of the ends at
the supports that would be inevitable with deflections encoun-
tered in conventionally reinforced concrete.
The ends of the plates behind the anchorage are thickened
to provide for distribution of the concentrated force applied
by the tendoms. Additional reinforcing in the form of trans-
verse and tie steel is provided in the area immediately behind
the tendon anchorages to resist the bursting of the concrete
A,
created by the ores tress force.
The end leaves of a transverse section through the
4
"Long span prestressed concrete folded platf* roofs", by
J. Brough and B. Stephens, Proc. km. 3oc. 0. *•« Jan. I960,
Vol. 85, p. 91.
5*
structure must be provided with tension ties to resist the
horizontal force component. This may be accomplished by pro-
viding a tie beam, gable wall or other means, at or near the
support, that is adequate for resistance of horizontal reac-
tion.
55
CONCLUSIONS
I'he first structures of this kind were large coal bunkers
designed and erected by G. Ehlers of Germany in 1924—25. The
first paper on the subject was published by him in 1930. His
analysis assumed the longitudinal joints to be hinged, neglect-
ing the transverse moments at the junction of the plates. The
displacement of joints was also ignored. This theory was im-
proved upon in 1932 by £• Gruber who included the effects of
transverse continuity and joint displacements. Assuming the
joints to be hinged as a first approximation, he developed a
solution in the form of simultaneous differential equations of
the fourth order, which were solved by the use of rapidly con-
verging series. This approach involves (7n + 2) unknowns for
(n + 1) plates; thus, a roof of 5 plates would involve 30 un-
knowns. Although solution proposed by Gruber was very labori-
ous, his conclusion that the assumption of hinged or rigid
joints would considerably affect the final results was signifi-
cant. This work was followed by that of Cramer who published
a paper in 1953. He laid down rough limits in terms of the
length bo width ratio of individual plates for their classifi-
cation as "long" and "short". The paper by Winter and Pei pub-
lished in 194-7 is a landmark in the theory on che subject as
thi,/, for the first time, reduced the algebraic solution into
a stress distribution procedure analogous to the well-known mo-
ment distribution method. However, they neglected the displace*
ment of the joints. For short folded plates their approach
56
offers a very simple desig,_i procedure. However, for long
plates the joint deflections cannot be ignored, Girkmann, in
his book published in 194-8, takes into account joint displace-
ments. Treating transverse moments at the joints as the un-
knowns, he formulated conditions for the compatibility of lon-
gitudinal stresses and displacements at joints. The method
leads to as many simultaneous equations as the unknown trans-
verse moments. The paper presented by Whitney at the joint
ASCE-IABSE meeting in New York is a presentation in English of
the Girkmann method with some modification. Gaafar in 1953
published a modification of the Winter and Pie method extended
to include the effect of joint displacements.
Among available methods, the winter and i ; ie procedure is
the simplest. It is applicable only to short folded plates for
which the joint displacements can be ignored without appreciable
error. Of the methods that are applicable to folded plates of
all proportions, those due to Gaafar which is considered in this
report seems to be most suitable.
57
ACKNOWLEDGMENT
The writer wishes to express his sincere gratitude to
Dr. John Mc Entyre for his kind guidance and assistance in
the preparation of this report.
58
APPENDIX I - EXPLANATION OF TERMS
a ■ Slope rise on 12
A = Area of concrete
c
A ■ Area of steel
e.g. = Center of gravity
dl ■ Dead load
e = Ecentricity of prestress force from the e.g.
e' = Distance from surface to center of prestress force
F ■ Prestress force after loboes
F ■ Modulus of rupture of concrete
F^ = Prestress force at tiiae of transfer
o
f = Allowable concrete unit stress
c
f ■ ■ Design concrete strength
f ■ Allowable unit stress
s
H ■ Vertical height of section valley to ridge
h * Vertical distance above or below e.g. of concrete
section
I ■ Moment of inertia of section
i ■ Subscript indicating condition at transfer of prestress
k = Kern point
11 = Live load
M m Bending moment
Q ■ Statical moment about e.g.
r = Radius of gyration
S = Principal tensile stress
S' = Principal tensile stress for ultimate
59
S. * Principal tensile stress
S. , » Ultimate principal tensile stress
c
t = Thiclmess of slab normal to section
t' * Thickness of slab vertically
tl - Total load
v = Unit 3 hear
v » Unit shear stress in concrete
c
v = Ultimata shear stress
V « Ultimate concrete shear
c
60
APPENDIX II - BIBLIOGRAPHY
"Design of Folded Plate Roofs," by Howard Simpson,
Proc. Am. Soc. C. E. , Vol. 84, No. STI, January, 1958.
"Hipped Plate Construction," by G. Winter and M. Pie,
Jour. Am. Con. Inst., Proc. Vol. 43, January, 1947.
"Reinforced Concrete Folded Plates Construction," by
G. S. Whitney, B« G. Anderson, and N. Birnbaum, Jour. Struct.
Div. , Proc. Am. Soc. C. £•, Vol. 84, January, 1958.
"Hipped Plate Analysis, Considering Joint Displacements,"
by I. Gaafar, Tran. Am. Soc. C. £. , Vol. 119, 1954.
"Design of Folded Plates," by Eliahu Traum, Proc. Am.
Soc. C. X., Vol. 85, No. STI. 8., October, 1959.
"Long Span Prestressed Concrete Folded Plate Roofs," by
J. Brough and B. Stephens, Proc. Am. Soc. C. E. , Vol. 119,
1954.
"Design of Prestressed Concrete Structures," by T. Y. Lin.
"Tentative Recommendations for Prestresaed Concrete," by
ACI-ASCE Joint Committee 325.
"Design and Calculation of Reinforced Concrete," by
K. L. Rao.
"The Analysis and Design of Folded Plates," by Ramswamy,
Ramaiah and Jain, Indian Concrete Journal, Jul} , 1961.
ANALYSIS OF PRESTRESSED AND REINFORCED
CONCRETE FOLDED PLATES
by
RAMESH GAKI
B. S., Bombay University, 1958
AN ABSTRACT OF
A MASTER'S REPORT
submitted in partial fulfillment of the
requirements for the degree
MASTER OF SCIENCE
Department of Civil Engineering
KANSAS STATE UNIVERSITY
Manhattan, Kansas
1963
Approved by:
Mix*
ajor Professor
The simplified methods of analysis for lor.p. spun pro-
stressed and reinforced concrete folded plr>te structures are
presented herein,
A reinforced concrete folded plate is analyzed b,/ solving
the applied load into two directions, one vertical and the
other parallel to the plate on which it acts. Vertical load
will induce "slab" action and the parallel load will induce
"plate" action, olab action can be analyzed by assuming non-
yielding supports and then applying a correction for deflec-
tion, from which total loads are found, 1 ,e" action causes
a deflection of the longitudinal edge and introduces longitu-
dinal stresr.es. To satisfy the compatibility condition a
stress distribution is done, homents are computed due to rela-
tive displacements. Prom these moments plate loads and longi-
tudinal stresses are calculated. This process is repeatedly
carried out and eve y time correction is applied. This correc-
tion will be in form of geometric series, 3ummation of this
series gives the actual deflection. The ratio of actual de-
flection and initial deflection gives the multiplier which,
when multiplied with initial lon ; ituainal stresses, ^ives ac-
tual longitudinal stresses. The final longitudinal stress is
calculated by adding stresses due to loading and stresses due
to deflections. tmantl are calculated at quarter point, end,
and at half distance of the longitudinal edge.
To analyze the prestressed plate, the stress distribution
along the longitudinal axis i3 analyzed in bhe same manner as
for any homogenous beam of rectangular section.
Transverse bending in successive plates is analyzed by
the conventional moment distribution method utilized for con-
tinuous structures.
A typical problem is outlined and the simplified method
o- analysis is discussed in this report.